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Prove that there does not exist any function $f:\mathbb{R}\to\mathbb{R}$ such that $$f(f(x))=x^2-2016$$ for all real number $x$.

My attempt :

Substitute $x=f(x)$ in $f(f(x))=x^2-2016$,

we have $f(f(f(x)))=f(x)^2-2016$

so $f(x^2-2016) = f(x)^2-2016$ ---[1]

$f$ is onto $(-2016, \infty)$

Since $x^2 \geq 0$ so $x^2-2016 \geq -2016$

$f(f(x))\geq -2016$

substitute $x=-x$ in[1], $f(x)^2-2016=f(-x)^2-2016$

so $f(x)^2 =f(-x)^2$

Please suggest how to proceed.

leonbloy
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user403160
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  • You sure that $ f(f(f(x))) = f(x^2) + c $ ? $ f(f(f(x))) $ should be equal to $ f(x^2 +k) $. Note that f(f(x)) = f o f (x) juste composing twice the same function on x... – user577215664 Sep 02 '17 at 04:11
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    There are answers to the question $f(f(x)) = x^2 - 2$ on this page. See if you understand any of them. Seems like the question is more complicated than it looks. https://math.stackexchange.com/questions/481017/find-fx-such-that-ffx-x2-2 – Sarvesh Ravichandran Iyer Sep 02 '17 at 04:25
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    @Isham, I've added the detail. – user403160 Sep 02 '17 at 04:25
  • @астон вілла олоф мэллбэрг. Thank you. I wonder if there will be elementary solution to this problem. – user403160 Sep 02 '17 at 04:39
  • @carat I wouldn't think so. For one, when you say that your function can do what it wants, which is a lack of regularity, you even bring pathological examples into play, and that's why the question is a lot harder. – Sarvesh Ravichandran Iyer Sep 02 '17 at 04:41
  • Thank you all. I found another helpful link, see problem 7 here : http://www.imomath.com/index.php?options=341&lmm=0 – user403160 Sep 02 '17 at 05:17

1 Answers1

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By the answer linked in the comments, no such function exists. Using the notation at the linked MSE response, you would have $\Delta(f) = 0 - 4(1)(-2016) > 1$, which rules out existence of a function satisfying the desired criterion (cf. Theorem 6 in the following pdf).

Benjamin Dickman
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